Question 1

What is the linear function of f(2/3)=-15/2 and f(-4)=-11

Question 2

f(2/3)=-15/2, x = 2/3, y = -15/2

f(-4)=-11, x = -4, y = -11.

we'll do the same here, so let's proceed,

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ (2)/(3) &,& -(15)/(2)~) % (c,d) &&(~ -4 &,& -11~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-11-\left(-(15)/(2) \right)}{-4-(2)/(3)}\implies \cfrac{-11+(15)/(2) }{-4-(2)/(3)}$

$\bf \cfrac{\quad (-7)/(2)\quad }{-(14)/(3)}\implies -\cfrac{7}{2}\cdot -\cfrac{3}{14}\implies -\cfrac{1}{2}\cdot -\cfrac{3}{2}\implies \cfrac{3}{4} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-\left( -\cfrac{15}{2} \right)=\cfrac{3}{4}\left(x-\cfrac{2}{3}\right) \\\\\\ y+\cfrac{15}{2}=\cfrac{3}{4}x-\cfrac{1}{2}\implies y=\cfrac{3}{4}x-\cfrac{16}{2}\implies y=\cfrac{3}{4}x-8$

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